The vertices of a hyperbola H are (± 6, 0) and its eccentricity is...

Given equation of hyperbola is

$H: \dfrac{{x}^{2}}{36} - \dfrac{{y}^{2}}{9} = 1$

Any point on the hyperbola is $P\left(6sec\theta, 3tan\theta\right)$.

Equation of normal is

$6xcos\theta + 3ycot\theta = 45$

Also, this normal is parallel to the line $\sqrt{2}x + y = 2\sqrt{2}$, therefore slope of both lines must be equal , hence

$- 2sin\theta = - \sqrt{2}$

$⇒ sin\theta = \dfrac{1}{\sqrt{2}}$

$⇒ \theta = \dfrac{\pi}{4}$

Equation of normal is

$\sqrt{2}x + y = 15$

And, this normal intersects $y -$axis at $K$, so

$P ≡$\left(6\sqrt{2}, 3\right)$and K ≡ \left(0, 15\right)$.

So,

$P{K}^{2} = {d}^{2} = {\left(6\sqrt{2} - 0\right)}^{2} + {\left(3 - 15\right)}^{2}$

$⇒ {d}^{2} = 216$